What is the volume at `N.T.P.` of a gas that occupies `43.0 mL` at `- 3^(@)C` and `0.98 "bar"` ?

Explanation

`V_(1) = 43.0 mL , V_(2) = ?`
`P_(1) = 0.98 "bar" , V_(2) = ?`
`T_(1) = -3 + 273 = 270 K , T_(2) = 273 K`
According to ideal gas equation.
`(P_(1)V_(1))/(T_(1)) = (P_(2)V_(2))/(T_(2))` or `V_(2) = (P_(1)V_(1)T_(2))/(T_(1).P_(2)) = ((0.98 "bar" ) xx (43 mL) xx (273 K))/((270 K) xx (1.013 "bar")) = 42.06 mL`